Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following got no answer on mathstackexchange. I believe it not to be hard, but maybe it is a little specialized?

All varieties will be over $\mathbb{C}$ and projective unless stated otherwise.

In Beauville - complex algebraic surfaces, the following is described: Let $S$ be a smooth surface and $p \in S$ a point. Let $\epsilon: \tilde S \rightarrow S$ be the blowup at $p$ and $E$ the resulting exceptional curve. Then $$ \text{Pic}(\tilde S) \cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} E $$ With $\text{Pic}$ i mean either the group of invertible sheaves or those of Cartier divisors modulo equivalence.

Question 1: I was wondering about the situation when $p$ is a simple double singularity, a node, instead of being smooth and the rest of $S$ is smooth. Does the same formula hold? If not, is there is a similar formula that describes $\text{Pic}(S)$ as a direct summand of $\text{Pic}(\tilde S)$?

Question 2: Does anybody know a reference for this situation: relationship of Picard groups of singular surfaces (or varieties in general) with their smoothification?

My guess and thoughts so far:

My first guess was that the same would hold, but i think that is false. The question is treated locally in Hartshorne example 6.5.2, which examines $$ \text{Spec}(\mathbb{C}[x,y,z]/(xy - z^2)) $$ The Weil class group of this affine variety is $\mathbb{Z}/2\mathbb{Z}$ and is generated by a ruling of the cone. This ruling is not a Cartier divisor and the cartier divisor class group is trivial.

This makes me conjecture: $$ \text{WCl}(\tilde S) \cong \epsilon^* \text{Wcl}(S) \oplus \mathbb{Z} E $$ $$ \text{Pic}(\tilde S) \cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} R \oplus \mathbb{Z} E $$ where $R$ corresponds to a ruling of the cone, which was a weil divisor on the singular variety but after the blowup corresponds to a Cartier divisor as well. $\text{WCl}$ means the group of Weil divsors modulo equivalence.

This is just an intuitive guess coming from Hartshorne's example so please please correct me if i'm wrong.

Thanks!

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

This is a partial answer to the part concerning Weil divisors.

I prefer to use the terminology of cycles of codimension $1$ instead of Weil divisors. The group of Weil divisors modulo rational equivalence is the Chow group $A^1$.

Let $f : \tilde{S}\to S$ be any proper birational morphism of integral normal algebraic varieties (integral normal schemes of finite type over a field) of dimension $\ge 2$. Let $E_1, \dots, E_n$ be the irreducible components of codimension $1$ (in $\tilde{S}$) of the exceptional locus of $f$.

As $f$ is proper, there is a canonical pushforward map $f_\*: A^1(\tilde{S})\to A^1(S)$ (see Fulton, Intersection Theory, Chapter 1) and it is a group homomorphism. As $S$ is normal, $f$ is an isomorphism outside of a codimension $\ge 2$ closed subset of $S$. In particular, $f_\*$ is surjective because for any irreducible cycle $\Gamma$ of codimension $1$ in $S$, $f_*$ maps the class of the strict transform $\tilde{\Gamma}$ to the class of $\Gamma$. However, in general $f_\*$ doesn't have a section.

By construction, the kernel of $f_\*$ is generated by the classes of the $E_i$. Let us show the map $\oplus_{1\le i\le n} [E_i]\mathbb Z \to A^1(\tilde{S})$ is injective. Let $g\in k(\tilde{S})$ be a rational function with divisor $\mathrm{div}_{\tilde{S}}(g)$ supported in $\cup_i {E_i}$. Then the divisor $\mathrm{div}_{S}(g)$ of $g$ as a rational function on $S$ is $0$. As $S$ is normal, this forces $g$ to be a unit in the ring of regular functions $O(S)$, hence $g$ is a unit in $O(\tilde{S})$ and $\mathrm{div}_{\tilde{S}}(g)=0$. Conclusion, we have an exact sequence $$0\to \oplus_{1\le i\le n} [E_i]\mathbb Z \to A^1(\tilde{S}) \to A^1(S) \to 0.$$

share|improve this answer
add comment

This is a complement to Sándor's remarks about the Cartier divisor case.

The ultimate reference for rational singularities of surfaces is Lipman's paper in Pub. Math. IHES 1969, "Rational Singularities ...", from which you can extract the exact sequence

$0 \to \textrm{Pic} S \xrightarrow{\epsilon^*} \textrm{Pic} \tilde{S} \xrightarrow{\theta} E^*$

where $E$ is the subgroup of $\textrm{Pic} \tilde{S}$ generated by the exceptional curves, $E^*$ is $\textrm{Hom}(E,\mathbb{Z})$ and the map $\theta \colon \textrm{Pic} \tilde{S} \to E^*$ is the one induced by the intersection pairing.

Now, in the case of a straightforward $A_1$ singularity, there is only one exceptional divisor and it has self-intersection $-2$. So we have $E^* \cong \mathbb{Z}$ and $\theta(E)$ is of index $2$ in $E^*$. The question is then whether $\theta$ is surjective, that is, whether there is a divisor on $\tilde{S}$ having odd intersection number with the exceptional divisor. In the case of a cone, as you describe, the straight line $R$ has intersection number $1$ with the exceptional divisor and so $\theta$ is surjective. But it's also possible to have an $A_1$ singularity where there is no such divisor - in that case, $\theta$ is not surjective. This really depends on the Zariski-local structure of the singularity, not just the analytic- (or étale-) local structure.

Incidentally, you can extend the above exact sequence by adding $\dotsb \to \textrm{Br} S \to \textrm{Br} \tilde{S}$ on the right-hand side, so we get information about the Brauer group of $S$ as well.

share|improve this answer
add comment

Joachim, your intuition is right, you will not have that same formula in general. Let's stick to the case you describe in the end.

Indeed $R$ will be a Cartier divisor on $\bar S$ and hence gives an element of $\text{Pic}\\,\bar S$, but it is not in $\epsilon^*\text{Pic}\,S$. However, it is also not a direct summand, because $2R$ is a Cartier divisor on $S$ so it is in $\epsilon^*\text{Pic}\\,S$.

So a more reasonable conjecture is that $$ \text{Pic}\\,\tilde S \simeq G \oplus \mathbb{Z} E, $$ where $G$ is a $\mathbb Z/2\mathbb Z$ extension of $\epsilon^* \text{Pic}(S)$, that is, there exists a short exact sequence: $$ 0\to \epsilon^* \text{Pic}(S) \to G \to \mathbb Z/2\mathbb Z \to 0, $$ where $\mathbb Z/2\mathbb Z$ is generated by $R+\epsilon^* \text{Pic}(S)$.

I believe that essentially the same proof works.

As for your conjecture for the $\text{WCl}$ (which is usually denoted by $\text{Cl}$, see [Hartshorne]), the initial problem is that you can't actually define $\epsilon^*$ for Weil divisors. That is, you can come up with various definitions that will have various properties, but in general you run into trouble either with respecting linear equivalence or the group structure.

There is a way to pull back $\mathbb Q$-Cartier divisors at least numerically and in this case that would indeed give what you conjecture via the above.

As for references, you could check out the papers of de Fernex-Hacon, Urbinati, and Chiecchio. (Start with the last two on arXiv (or MArXiv.org), they have fewer papers and they reference the one by de Fernex-Hacon that's relevant).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.